Question

a set of art exam scores are normally distributed with a mean of 81 points and a standard deviation of 10 points. kamil got a score of 78 points on the exam. what proportion of exam scores are lower than kamils score? you may round your answer to four decimal places.

Explanation:

Step1: Calculate the z - score

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x$ is the value from the data set, $\mu$ is the mean, and $\sigma$ is the standard deviation. Given $\mu = 81$, $\sigma=10$, and $x = 78$. Then $z=\frac{78 - 81}{10}=\frac{-3}{10}=- 0.3$.

Step2: Find the proportion using the standard normal distribution

We want to find $P(X\lt78)$, which is equivalent to $P(Z\lt - 0.3)$ in the standard - normal distribution. Looking up the value of $P(Z\lt - 0.3)$ in the standard normal table (or using a calculator with a normal - distribution function), we get $P(Z\lt - 0.3)=0.3821$.

Answer:

$0.3821$